If 3.42 grams of sucrose, C12H22O11, are dissolved in 20.0 grams of water, what will be the boiling point of the resulting solution? Kb for water = 0.512 oC/m.
Answers
Answered by
3
Answer:
Let the molality of the solution is m.
ΔT
f
=K
f
×m andΔT
b
=K
f
b×m
ΔT
f
+ΔT
b
=(K
f
+K
b
)×m=(1.86+0.51)×m
- =2.37×m
Hencem=5/2.37=2.11
and
1000
2.11×100
=0.211 gram of sucrose must be dissolved in 100 gram of water.
Required mass of sucrose =0.211mol×342g/mole=72.2g
So option B is correct.
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